Separation of a point and a compact set ($x\notin K⟹x\in V,K\subseteq U,U\cap V=\varnothing$)

In a Hausdorff space, a point $x$ outside a compact subset $K$ can be separated from all of $K$ by disjoint open sets. Choose disjoint open neighbourhoods $U_k$ of $k$ and $V_k$ of $x$ for each $k\in K$. Finitely many $U_k$ cover $K$ by compactness. Their union and the matching finite intersection of the $V_k$ give the displayed separation. If $K$ is empty, choose $U=\varnothing$, $V=X$. Thus $X\setminus K$ is open, proving that $K$ is closed.

Two disjoint compact subsets of a Hausdorff space have disjoint open neighbourhoods. For every $p\in P$, separation of a point and a compact set gives disjoint open sets $G_p\ni p$ and $H_p\supseteq Q$. Choose finitely many $G_p$ covering $P$, then take their union for $U$ and the matching finite intersection of the $H_p$ for $V$. The compactness of both subsets is used, once in the point-versus-compact set result and once in the outer finite selection. The ambient Hausdorff space need not be compact or a normal space.